Let $f(x)=x^3+9x^2+13x$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $-7\leq x \leq-1$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-6$ (Choice B) B $-5$ (Choice C) C $-3$ (Choice D) D $-2$
Explanation: According to the Mean Value Theorem, there exists a number $c$ in the open interval $-7<x<-1$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(-1)-f(-7)}{(-1)-(-7)}$ First, let's find that average rate of change: $\dfrac{f(-1)-f(-7)}{(-1)-(-7)}=\dfrac{-5-7}{6}={-2}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={-2}$. $f'(x)=3x^2+18x+13$ The solutions of $f'(x)=-2$ are $x=-5$ and $x=-1$. Out of these, only $x=-5$ is within the interval $-7<x<-1$. In conclusion, $c=-5$.